A nucleus is a bound system. That is, you would need to supply energy to disperse the nucleons by breaking the nuclear bonds between them. FIGURE 42.5 shows this idea schematically.

FIGURE 42.5 The nuclear binding energy.

You learned a similar idea in atomic physics. The energy levels of the hydrogen atom are negative numbers because the bound system has less energy than a free proton and electron. The energy you must supply to an atom to remove an electron is called the ionization energy.

In much the same way, the energy you would need to supply to a nucleus to disassemble it into individual protons and neutrons is called the binding energy. Whereas ionization energies of atoms are only a few eV, the binding energies of nuclei are tens or hundreds of MeV, energies large enough that their mass equivalent is not negligible.

Consider a nucleus with mass $m_{\text{nuc}}$. It is found experimentally that $m_{\text{nuc}}$ is less than the total mass $Z{m}_{\text{p}}+N{m}_{\text{n}}$ of the $Z$ protons and $N$ neutrons that form the nucleus, where $m_{\text{p}}$ and $m_{\text{n}}$ are the masses of the proton and neutron. That is, the energy equivalent $m_{\text{nuc}}{c}^2$ of the nucleus is less than the energy equivalent $(Z{m}_{\text{p}}+N{m}_{\text{n}})c^2$ of the individual nucleons. The binding energy $B$ of the nucleus (not the entire atom) is defined as

This is the energy you would need to supply to disassemble the nucleus into its pieces.

The practical difficulty is that laboratory scientists use mass spectroscopy to measure atomic masses, not nuclear masses. The atomic mass $m_{\text{atom}}$ is $m_{\text{nuc}}$ plus the mass $Z{m}_{\text{e}}$ of $Z$ orbiting electrons. (Strictly speaking, we should allow for the binding energy of the electrons, but these binding energies are roughly a factor of ${10}^6$ smaller than the nuclear binding energies and can be neglected in all but the most precise measurements and calculations.)

Fortunately, we can switch from the nuclear mass to the atomic mass by the simple trick of both adding and subtracting $Z$ electron masses. We begin by writing Equation 42.4 in the equivalent form

Now $m_{\text{nuc}}+Z{m}_{\text{e}}=m_{\text{atom}}$, the atomic mass, and $Z{m}_{\text{p}}+Z{m}_{\text{e}}=Z(m_{\text{p}}+m_{\text{e}})=Z{m}_{\text{H}}$, where $m_{\text{H}}$ is the mass of a hydrogen atom. Finally, we use the conversion factor $1\text{u}=931.49\text{MeV}\text{/}{c}^2$ to write $c^2=931.49\text{MeV}\text{/}\text{u}$. The binding energy is then

where all three masses are in atomic mass units.

The nuclear binding energy increases as $A$ increases simply because there are more nuclear bonds. A more useful measure for comparing one nucleus to another is the quantity $B\text{/}A$, called the binding energy per nucleon. Iron, with $B=493\text{MeV}$ and $A=56$, has 8.80 MeV per nucleon. This is the amount of energy, on average, you would need to supply in order to remove one nucleon from the nucleus. Nuclei with larger values of $B\text{/}A$ are more tightly held together than nuclei with smaller values of $B\text{/}A$.

FIGURE 42.6 is a graph of the binding energy per nucleon versus mass number $A$. The line connecting the points is often called the curve of binding energy. This curve has three important features:

• There are peaks in the binding energy curve at $A=4\text{,}12$, and 16. The one at $A=4$, corresponding to ${}^4\text{He}$, is especially pronounced. As you’ll see, these peaks, which represent nuclei more tightly bound than their neighbors, are due to closed shells in much the same way that the graph of atomic ionization energies (see Figure 41.22) peaked for closed electron shells.

• The binding energy per nucleon is roughly constant at $\approx 8\text{MeV}$ per nucleon for $A>20$. This suggests that, as a nucleus grows, there comes a point where the nuclear bonds are saturated. Each nucleon interacts only with its nearest neighbors, the ones it’s actually touching. This, in turn, implies that the nuclear force is a short-range force.

• The curve has a broad maximum at $A\approx 60$. This will be important for our understanding of radioactivity. In principle, heavier nuclei could become more stable (more binding energy per nucleon) by breaking into smaller pieces. Lighter nuclei could become more stable by fusing together into larger nuclei. There may not always be a mechanism for such nuclear transformations to take place, but if there is a mechanism, it is energetically favorable for it to occur.

FIGURE 42.6 The curve of binding energy.